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The odd perfect number problem likely needs no introduction. Recent progress (where by recent I mean roughly the last two centuries) seems to have focused on providing restrictions on an odd perfect number which are increasingly difficult for it to satisfy (for example, congruence conditions, or bounding by below the number of distinct prime divisors it must have). By reducing the search space in this manner, and probably due to other algorithmic improvements (factoring, parallelizing, etc.), there has also been significant process improving lower bounds for the size of such a number. A link off of oddperfect.org claims to have completed the search up to $10^{1250}$.

But, assuming my admittedly cursory reading of the landscape is correct, none of the current research seems particularly equipped to prove non-existence. The only compelling argument I've seen on this front is "Pomerance's heuristic" (also described on oddperfect.org). Worse, and maybe this is really the point of this question, it would be a little disappointing if the non-existence proof was an upper bound of $10^{1250}$ (depending on the techniques used to get the bound) combined with the above brute force search.

On the other hand, maybe there's some hope that some insight can be gained into the sum-of-divisors function by modern techniques. For example, the values of the arithmetic functions $$ \sigma_{k}(n):=\sum_{d\mid n}d^k, $$ for $k\geq 3$ odd, arise as coefficients of normalied Eisenstein modular forms, and the study of said forms gives amazing proofs of amazing identities between them. For $k=1$, the case of interest, the normalized Eisenstein series $E_2$ is only "quasi-modular", but such forms satisfy sufficiently nice transformation properties that I wonder if $E_2$ has anything to say about the problem.

Since no doubt many people on this site will be able to immediately address the previous idea (so please do!), my more general question is whether or not there are applications of the modern machinery of modular forms, mock modular forms, diophantine analysis, Galois representations, abc conjecture, etc., that have anything to say about the odd perfect number problem. Does it descend from or relate to any major open problems from modern algebraic/analytic number theory?

Aside: I hope this does not come off as dismissive of "elementary" techniques, or of the algorithmic ones mentioned in the first paragraph. Indeed, they have, to my knowledge, been the only source of progress on this problem, and certainly contain interesting mathematics. Rather, this phrasing stems from my desire to find anything in the intersection of "odd perfect number theory" and "things I know anything about," and perhaps a desire to see the odd perfect number problem settled without the use of a beyond-gigantic brute force search.

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Cam, in the title it should read algebraic, although it is hard to imagine anyone misunderstanding. –  Will Jagy Sep 20 '10 at 20:53
    
Thanks, Will and Charles. –  Cam McLeman Sep 20 '10 at 23:23
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+1, really liked it. –  Unknown Sep 21 '10 at 8:53

6 Answers 6

This is a problem I have thought alot about. I have not seen any of the modern techniques in your list applied to the problem. Part of the issue is that if you represent $\sigma(n)=2n$ as a Diophantine equations in $k$ variables (corresponding to the prime factors--but allowing the powers to vary) then there are lots of solutions (just not where all the variables are simultaneously prime). So the usual methods of trying to show non-existence of solutions just don't cut it. Historically, this multiplicative approach is the one many people have taken, because at least some progress can be made on the problem. My personal feeling is that maybe someday these bounding computations will be tweaked to the point that they lead to the discovery of some principle that will solve the problem. For example, in one of my recent papers, I was led to consider the gcd of $a^m-1$ and $b^n-1$ (where $a$ and $b$ are distinct primes). I would conjecture that this gcd has small prime factors unless $m$ or $n$ is huge. If that happens, many of the computations related to bounding OPNs become much easier.

I have occasionally thought about whether modular forms might say something about this topic (which is why I'm currently sitting in on my colleague's course). Instead of $\sigma(n)$, the `right' function to consider is $\sigma_{-1}(n)=\sigma(n)/n$ and I don't know off the top of my head if it appears in connection with (weakly holomorphic) modular forms. But I know there are some nice techniques about multiplicative functions that decrease over the primes, etc...

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Thanks for this response. Can you comment any further on sigma_{-1}, and why you think it's 'right'? –  Cam McLeman Sep 21 '10 at 12:47
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I think it is right for a few reasons. First, because the problem then becomes $\sigma_{-1}(n)=2$, so we are simply asking if $\sigma_{-1}$ ever represents $2$ on odd input. Perhaps there is some general theory that multiplicative functions satisfying growth conditions like $\sigma_{-1}$ only represent numbers finitely many times. Something like that would be nice! Second, $\sigma_{-1}$ makes sense as you allow prime powers to approach infinity--which allows a few more techniques to come into play. More multiplicative structure is available ($\sigma_{-1}$ decreases as primes increase). –  Pace Nielsen Sep 21 '10 at 15:18
    
(I seem to recall a conversation about another multiplicative function that represented numbers only finitely many times on odd input. It was related to modular forms, but I forget which function it was.) –  Pace Nielsen Sep 21 '10 at 15:21
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Cam, a quick comment about something someone recently pointed out to me. While it is true that $E_{2}$ is not modular, it does turn out that the odd terms of $E_{2}$ are modular of level 4. And of course, the odd terms are all you need to consider when looking for OPNs. –  Pace Nielsen Dec 7 '10 at 17:53

For links between perfect numbers and the ABC conjecture, see this paper by Luca and Pomerance.

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Thanks. I did at one point look through there quickly for references to odd perfect numbers -- there didn't seem to be much. Granted, I haven't digested the core of the material there, so there may be more links that didn't explicitly get mentioned. –  Cam McLeman Sep 23 '10 at 18:50

I had an opportunity to work with Dr. Beauregard Stubblefield 35 years ago (yep, 62 and proud of it) and he generously gave the credit for the group's result to Dr. Mary Buxton and me. Dr. Stubblefield of course did most of the work, and he and I discussed many things. It seems that Leopold Kronecker, the great German algebraist 1823-1891, had it right the whole time. He proved that X = p^e + p^(e+1) + ... + p^2 + p + 1, where e is one less than a prime and p is a prime, cannot be algebraically reduced. But we knew that. The big deal is that the numeric factors of the expression are either (e + 1) || X or (k(e + 1) + 1) | X. Dr. Stubblefield found the pattern and I found that Kronecker had proved it. Stubblefield used the result in his Proposition 11, which he proved. Thirty-five years ago we used the result to factor many sigma(p^e). We talked about extending the result back then so we both had deep input into the new theory. Recently, after an engineering career in the auto industry in Detroit, I extended Proposition 11 to apply to many more cases in several different ways. I think, Cam, that this is the road you are suggesting. Steve Elmore

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I agree with Pace - the correct function to consider would be the abundancy index instead of the sigma function itself. In a certain sense, the abundancy index value of 2 for perfect numbers (odd or even) has served as a baseline on which various other properties and concepts related to number perfection were developed.

Additionally, although Balth. Van der Pol was able to derive the congruence conditions mentioned, for an OPN N (which effectively gave us two cases: case (1) $3 | N$, case (2) $N$ is not divisible by $3$) using a recursion relation satisfied by the sigma function (and this was achieved via a nonlinear partial differential equation), the same result follows from eliminating the case $N \equiv 2\pmod 3$ and then using the Chinese Remainder Theorem afterwards.

Though I have seen modular forms before (having done a [modest] exposition of elliptic curve theory and related topics) for my undergraduate thesis, I have likewise not seen any 'objects' used in Wiles' FLT proof applied directly to the OPN conjecture -- at least, not 'directly' in the literal sense of the word.

Just some thoughts about OPNs, now that you've mentioned them:

A closer look at the (multiplicative) forms of even and odd perfect numbers gives you:

Even PN = ${2^{p - 1}}(2^p - 1)$ = (even power of a "small" prime) x ("big" prime)

Odd PN = ${m^2}{q^k}$ = (even power of several primes) x ("another" prime "power")

Ronald Sorli conjectured in his Ph. D. thesis (titled Algorithms in the Study of Multiperfect and Odd Perfect Numbers, completed in 2003) that for an OPN, it was in fact (numerically) plausible that $k = 1$. (Compare that with the Mersenne prime. Incidentally, we call $q$ the Euler prime!)

If Sorli's conjecture is proven true, we will have (considerably) more information about an OPN's structure that could enable a quick resolution about its conjectured nonexistence.

Let me know if you need more information.

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I have always believed that there are no OPNs. In fact, I had been working on that problem since the year 2000. While I may or may not succeed, I certainly know others will (eventually, and hopefully before the year 2011 ends). Indeed, the three crucial points are to prove: (1) $I(p^k) < I(m)$ where $I(x)$ is the abundancy index of $x$. (2) Since prime powers are deficient, either $p < m$ or $m < p$. (3) Show that $k=1$ follows from (1) and (2). Good luck to all! –  Jose Arnaldo Dris Jan 29 '11 at 1:11
    
Update: Preprint on Sorli's conjecture has been released - <scribd.com/doc/46766721/OPNPaper3>;. –  Jose Arnaldo Dris Jan 29 '11 at 1:14

The odd perfect number problem does have a connection to modular forms. the divisor funct can be written as a function of the tau function and sigma_{k}(n) = sum_{d|n} d^k. The earlier example is the van der Pol identity. This was used by Touchard to conclude that n = 36a + 9 or 12b + 1.

The modular for should be normalized to use sigma(n)/n = 2 to see if it gives a contradiction.

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I don't understand. The divisor function is a sigma function, with k=1. –  Cam McLeman Sep 23 '10 at 18:54
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Cam, I think he means that it can be written in terms of higher sigma functions. I've seen Touchard's result, but that kind of conclusion follows much more quickly from considering congruence conditions, rather than complicated arguments. –  Pace Nielsen Sep 23 '10 at 22:34
    
Alternatively, we can try a (re)normalization using (say) $J(n) = \frac{1}{2}I(n)$, where $I(n)$ is the abundancy index of $n$. We just need to take note, though, that $I$ being multiplicative carries over to $J$ satisfying $J(xy) = 2J(x)J(y)$ for all $x$ and $y$ with $gcd(x, y) = 1$. –  Jose Arnaldo Dris Dec 5 '10 at 6:39

In fact, it might be possible to prove that there is no OPN provided one could prove that if $1+p+...p^k=q^n$, where $p$ and $q$ are odd primes and $k\ge 2$, then $p\lt q$. The trick is to write an OPN $m$ as $p_1^{e_1}...p_k^{e_k}$ and to consider the map $\tilde{\sigma}$ which maps each $p_{i}$ to $R(\sigma(p_i^{e_i}))$, with $R(n)$ the product of the primes dividing $n$.

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Indeed $\tilde{\sigma}$ should act as a permutation without fixed point on a subset $A$ of $\{p_1,...p_k\}$ (cause $\sigma$ is a multiplicative function). But if really $p\lt q$ in the first equation of my answer, then for each element $a$ of $A$, $a\lt \tilde{\sigma(a)}$. So that $\tilde{\sigma}$ can't be a permutation. Hence a contradiction, and this would prove that there is no OPN. –  Sylvain JULIEN Aug 23 '12 at 12:10
    
Unfortunately it does not hold. I forget the details, but perhaps a quick search among primes less than 100 and k=2 will provide a counterexample. Gerhard "Or Primes Less Than Ten" Paseman, 2012.08.23 –  Gerhard Paseman Aug 23 '12 at 18:28
    
In fact, 93*49 = 1 + 67*68. The example involving 7 and k=3 factors into 25 and 16. Gerhard "Ask Me About System Design" Paseman, 2012.08.23 –  Gerhard Paseman Aug 23 '12 at 18:35
    
And while the above examples do not directly address your question, to me they provide evidence that potentially p>q. Gerhard "Yes, Even For Odd Primes" Paseman, 2012.08.23 –  Gerhard Paseman Aug 23 '12 at 18:39
    
You might have misunderstood what I meant. What I actually meant is that whenever the conditions below simultaneously hold: a) $1+p+...p^{k}=q^n$ b) $p$ and $q$ are odd primes c) $k\geq 2$ then $p\lt q$ (or, equivalently, $k\gt n$). Do you agree that this would imply that there is no OPN? –  Sylvain JULIEN Aug 23 '12 at 19:57

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